(A) Elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$. For $NaCl$ $(58.4 \ g \approx 1 \text{ mole})$, $i = 2$. For gluc

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$NaCl$ solution will show higher elevation of boiling point

(A) Elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$. For $NaCl$ $(58.4 \ g \approx 1 \text{ mole})$, $i = 2$. For glucose $(180 \ g = 1 \text{ mole})$, $i = 1$. Since both solutes are dissolved in the same amount of solvent ($1000 \ mL \approx 1 \text{ kg}$ of water), their molality $(m)$ is the same. Therefore, $\Delta T_{b} \propto i$. Since $i_{NaCl} > i_{glucose}$, the $NaCl$ solution will show a higher elevation of boiling point.

Class 12 Chemistry Solutions Elevation of boiling point of the solvent

Difficult

$58.4 \ g$ of $NaCl$ and $180 \ g$ of glucose were separately dissolved in $1000 \ mL$ of water. Identify the correct statement regarding the elevation of boiling point $(b.p.)$ of the resulting solutions.

WBJEE 2012 All WBJEE

A
$NaCl$ solution will show higher elevation of boiling point
B
Glucose solution will show higher elevation of boiling point
C
Both the solutions will show equal elevation of boiling point
D
The boiling point elevation will be shown by neither of the solutions

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solutions

Subtopic

Elevation of boiling point of the solvent

Previous Year

WBJEE 2012 Paper All WBJEE years →

To observe an elevation of boiling point of $0.05^{\circ} C$, the amount of a solute (mol. wt. $= 100$) to be added to $100 \ g$ of water $(K_{b} = 0.5)$ is (in $g$)

MediumWBJEE 2014

View Solution

When $3.00 \, g$ of a substance $'X'$ is dissolved in $100 \, g$ of $CCl_4$,it raises the boiling point by $0.60 \, K$. The molar mass of the substance $'X'$ is $..... \, g \, mol^{-1}$. (Nearest integer).
$[$ Given $K_b$ for $CCl_4$ is $5.0 \, K \, kg \, mol^{-1} ]$

EasyJEE MAIN 2021

View Solution

Arrange the following solutions in order of their increasing boiling points.
$(i) \ 10^{-4} \ M \ NaCl$ $(ii) \ 10^{-4} \ M \ \text{Urea}$ $(iii) \ 10^{-3} \ M \ NaCl$ $(iv) \ 10^{-2} \ M \ NaCl$

MediumJEE MAIN 2025

View Solution

When a substance is dissolved in a solvent,the vapour pressure of the solvent is decreased. This results in:

Easy

View Solution

$A$ solution containing $0.1050 \ g$ of a substance in $15.84 \ g$ of ether has a boiling point $1 \ ^\circ C$ higher than that of pure ether. What is the molar mass of the substance? [Given: Molar elevation constant for ether per $100 \ g = 21.6$]

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$58.4 \ g$ of $NaCl$ and $180 \ g$ of glucose were separately dissolved in $1000 \ mL$ of water. Identify the correct statement regarding the elevation of boiling point $(b.p.)$ of the resulting solutions.

Similar Questions

To observe an elevation of boiling point of $0.05^{\circ} C$, the amount of a solute (mol. wt. $= 100$) to be added to $100 \ g$ of water $(K_{b} = 0.5)$ is (in $g$)

When $3.00 \, g$ of a substance $'X'$ is dissolved in $100 \, g$ of $CCl_4$,it raises the boiling point by $0.60 \, K$. The molar mass of the substance $'X'$ is $..... \, g \, mol^{-1}$. (Nearest integer).$[$ Given $K_b$ for $CCl_4$ is $5.0 \, K \, kg \, mol^{-1} ]$

Arrange the following solutions in order of their increasing boiling points.$(i) \ 10^{-4} \ M \ NaCl$ $(ii) \ 10^{-4} \ M \ \text{Urea}$ $(iii) \ 10^{-3} \ M \ NaCl$ $(iv) \ 10^{-2} \ M \ NaCl$

When a substance is dissolved in a solvent,the vapour pressure of the solvent is decreased. This results in:

$A$ solution containing $0.1050 \ g$ of a substance in $15.84 \ g$ of ether has a boiling point $1 \ ^\circ C$ higher than that of pure ether. What is the molar mass of the substance? [Given: Molar elevation constant for ether per $100 \ g = 21.6$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module

When $3.00 \, g$ of a substance $'X'$ is dissolved in $100 \, g$ of $CCl_4$,it raises the boiling point by $0.60 \, K$. The molar mass of the substance $'X'$ is $..... \, g \, mol^{-1}$. (Nearest integer).
$[$ Given $K_b$ for $CCl_4$ is $5.0 \, K \, kg \, mol^{-1} ]$

Arrange the following solutions in order of their increasing boiling points.
$(i) \ 10^{-4} \ M \ NaCl$ $(ii) \ 10^{-4} \ M \ \text{Urea}$ $(iii) \ 10^{-3} \ M \ NaCl$ $(iv) \ 10^{-2} \ M \ NaCl$